JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    To neutralize 25 ml of 0.25 M \[N{{a}_{2}}C{{O}_{3}}\] solution how much volume of 0.5 M \[HCl\] is required [MP PET 1994]

    A) 12.5 ml

    B) 25 ml

    C) 37.5 ml

    D) 50 ml

    Correct Answer: A

    Solution :

    \[{{M}_{1}}{{V}_{1}}\] =  \[{{M}_{2}}{{V}_{2}}\] \[(N{{a}_{2}}C{{O}_{3}})\]     =  \[(HCl)\] \[0.25M\times 25\]   =  \[0.5M\times {{V}_{2}}\] \[{{V}_{2}}=\frac{0.25M\times 25}{0.5M}=12.5\,\,ml\]


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